Graphing Linear Inequalities

Now that you have mastered how to graph one and two variable linear equations:

x + y = 3

It is time now to learn how to graph linear inequalities?

x + y is less than or equal to 3

 

First step: Graph the inequality as if were and equation

the graph of x + y is less than or equal to 3, which passes through the points ( 0 , 3 ) and ( 3 , 0 )
This takes care of the
x + y = 3 part …

So, what about the

<?

Remember that this rectangular coordinate grid is just made up of a bunch of ( x , y ) points.

All the points ON the line work in this part:

x + y = 3

Now, we just need to find all the points that work in this part:

x + y < 3

Here’s how we find these points:

We just pick a point OFF the line and test it!

the graph of x + y < 3 ... test the point ( 0 , 0 ) The origin, ( 0 , 0 ), is really easy to work with, so let’s try it:

x + y < 3 ... 0 + 0 < 3 ... 0 < 3

( 0 , 0 ) works in x + y < 3 !

Here’s the deal: 

If ( 0 , 0 ) works, then ALL the points
on the same side of the line will work.

Don’t believe me?  Let’s try a couple more:

the graph of x + y < 3 ... test the points ( -1 , 2 ) or ( 2 , 3 ) ... try plugging the point ( -1 , 2 ) into the equation x + y < 3 ... -1 + 2 < 3 ... 1 < 3 ... yep!  ... try plugging the point ( 2 , 3 ) into the equation x + y < 3 ... 2 + 3 < 3 ... 5 < 3 ...nope!

So, our answer is this:

a graph of x + y is less than or equal to 3 ... the portion of the graph under the line is shaded All the ( x , y ) points on the line and in the shaded region work in

x + y is less than or equal to 3

 

EXAMPLE :

Graph  2x + y > 4
First, do the line:the graph of the line 2x + y > 4 ... remember the line is dashed We make it a dashed line since there is no =.So, which side do we shade?Let’s try ( 0 , 0 ):

2x + y > 4 ... 2 ( 0 ) + 0 > 4 ... 0 > 4 ... Nope!

( 0 , 0 ) doesn’t work!  That means nothing on that side will work … So, all the good points must be on the other side!

the graph of 2x + y > 4 ... the portion of the graph above the line is shaded All the points in theshaded region work in

2x + y > 4

EXAMPLE 3:

Graph    y > ( 1 / 3 )x

* Remember — to graph this guy use y = mx + b.  And we’ll use a dashed line since there is no =.

the graph of y > ( 1 / 3 )x, which passes through the points ( -3 , - 1 ) , ( 0 , 0 ) , and ( 3 , 1 ) ... remember the line is dashed What side do we shade?The only glitch on this one is that we can’t try
( 0 , 0 )
because it’s ON the line.
Let’s try ( 1 , 2 ) :

y > ( 1 / 3 ) x ... 2 > ( 1 / 3 ) ( 1 ) ... 2 > ( 1 / 3 )

the graph of the line y = ( 1 / 3 )x ... test the point ( 1 , 2 )

Yep!  It works!  So, shade that side:

the graph of y > ( 1 / 3 )x ... the portion of the graph above the line is shaded All the ( x , y ) pointsin the shaded regionwork in

y > ( 1 / 3 )x

NOVEMBER 9 CLASSWORK

nov 9 alg classwork

NOVEMBER 9 CLASSWORK ANSWERS

nov 9 alg classwork answers

 

NOVEMBER 10 CLASSWORK

nov 10 graphing classwork

nov 10 translate inequalities

 

NOVEMBER 10 CLASSWORK ANSWERS

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nov 10 alg classwork answers

 

NOVEMBER 12 CLASSWORK

nov 12 alg classwork

NOVEMBER 12 CLASSWORK ANSWERS

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NOVEMBER 13 QUIZ

nov 13 alg quiz

 

MIDTERM REVIEW THIS WEEK

Midterm is this Friday, November 6.

midterm2

midterm

midterm4midterm5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

DAY 1 REVIEW – POWERS and EXPONENTS

exponent

 

 

 

 

 

 

exponent3

 

 

 

 

 

 

 

 

 

NOVEMBER 5 ALGEBRA WORD PROBLEMS

nov 5 alg classwork

 

OCTOBER 26 ALGEBRA CLASSWORK

oct 26 alg classwork

OCTOBER 26 CLASSWORK ANSWERS

oct 26 alg classwork answers

OCTOBER 28 ALGEBRA CLASSWORK

oct 28 alg exponent classwork

OCTOBER 28 ALGEBRA CLASSWORK ANSWERS

oct 28 alg exponent classwork answers

Simplifying Radicals

They created a fun little game called “rationalizing the denominator.”

Here’s how it goes:

3 / sqrt(2) = ( 3 / sqrt(2) )( sqrt(2) / sqrt(2) ) = ( 3*sqrt(2) ) / ( sqrt(2)*sqrt(2) ) = ( 3*sqrt(2) / sqrt(4) ) = ( 3*sqrt(2) ) / 2
Multiply by  1. 

ANOTHER EXAMPLE

5 / sqrt(10) = ( 5 / sqrt(10) )( sqrt(10)*sqrt(10) ) = ( 5*sqrt(10) / sqrt(100) ) = ( 5*sqrt(10) / 10 ) = sqrt(10) / 2     reduce 5/10 = 1/2  

 

What if we needed to get the radical out of this denominator?

5 / ( 2 - sqrt(3) )

Same game — we still multiply by 1…

06-Radicals-08.gif
The  1 will be made up of this guy’s conjugate.

Well now, that was helpful wasn’t it?

Conjugate = same guy, different sign

So, the conjugate of

2 - sqrt(3)   is   2 + sqrt(3)

Check out how it works:

( 5 / ( 2 - sqrt(3) )( ( 2 + sqrt(3) ) / ( 2 + sqrt(3) )
Put ( ) around everyone or you’re
really going to mess things up!

( 5 / ( 2 - sqrt(3) )( ( 2 + sqrt(3) ) / ( 2 + sqrt(3) ) = ( 10 + 5*sqrt(3) ) / ( 4 - sqrt(9) ) = ( 10 + 5*sqrt(3) ) / (4 - 3 ) = 10 + 5*sqrt(3)

Not only did we ditch the radical in the denominator,
in this one, we ditched the whole denominator.

 Another example:

4 / ( 1 + sqrt(7) ) = ( 4 / ( 1 + sqrt(7) ) )( ( 1 - sqrt(7) ) / ( 1 - sqrt(7) ) ) = ( 4 - 4*sqrt(7) ) / ( 1 - sqrt(49) ) = ( 4 - 4*sqrt(7) ) / ( 1 - 7 ) = ( 4 - 4*sqrt(7) ) / ( -6 ) = ( -2( -2 + 2*sqrt(7) ) ) / ( -6 ) = ( -2 + 2*sqrt(7) ) / 3

                      

 NOVEMBER 3 CLASSWORK

nov 3 alg classwork

NOVEMBER 3 CLASSWORK ANSWERS

nov 3 alg classwork answers

n0v 3 alg classwork A answers

NOVEMBER 4 CLASSWORK

nov 4 alg classwork

 

NOVEMBER 4 CLASSWORK ANSWERS

nov 4 classwork answers

Fractional Exponents

Nothing to fear.  It’s just a different notation for what you’ve already been doing.

cube root of 8 can be written as  8^(1/3)
sqrt(15) can be written as  15^(1/2)

* Remember that the square root symbol is really the square root symbol with a little 2 on it …  We just assume the 2.

Not that bad, is it?

Here’s the general rule:

the nth root of a = a^(1/n)

This notation has some advantages because it will help you understand some things better.

Check it out:

sqrt(5) * sqrt(5) = ( 5^(1/2) ) * ( 5^(1/2) ) = 5^( 1/2 + 1/2 ) = 5^1 = 5

Now, we can show why this rule works:

( the nth root of a ) * ( the nth root of b ) = the nth root of ab

( the nth root of a ) * ( the nth root of b ) = ( a^(1/n) ) * ( b^(1/n) ) = ( ab )^(1/n) = the nth root of ab

OCTOBER 27 ALGEBRA MIDTERM EXAM REVIEW

oct 27 alg classwork

OCTOBER 27 MIDTERM EXAM REVIEW ANSWERS

 oct 27 alg classwork answers

Elimination Method (Strange Solutions)

Here are some examples of really strange answers you have get when you are solving two equations.
Let’s graph these two lines:  pierre ant
x + y = 1x + y = 4

a graph of the lines x + y = 1 and x + y = 4

What’s the solution?  They have to intersect to get a solution.  But, that’s never going to happen since these lines are parallel.  So, there is no (x, y) that is going to work in both equations.

When you are solving by elimination or substitution, you’re sometimes going to get a false statement like

0 = 8

For the best answer, you’ll want to write something like this:

These lines are parallel.
There is no solution.

_________________________________________________

The other strange thing that can happen is that the two lines can be lying right on top of each other — they are the same line.

Let’s graph these two lines: pierre ant

x + y = 3

-2x - 2y = -6

a graph of the lines x + y = 3 and -2x - 2y = -6

How will you know when this strange answer announces itself to you?

When you’re solving by elimination or substitution…  you’ll be going along…  and you’ll get this:

0 = 0

 

 

So, what does this mean for our answer? 

Every single (x, y) point that appears in one line appears in the other.  So, every x and y combination (domain and range) that works in one equation, works in the other equation.  And how many points are in a line?  That’s right, there’s an infinite number of solutions!  So how many points that work in one line work in the other?  An infinite number. 

You should write something like this for your answer:

These are the same line, so there are
an infinite number of solutions.


The following table will guide us to choose the best method to solve.
Methods
When to choose
Graphing
When we are asked to estimate the solution. Graphing sometimes does not give the exact solution.
Substitution
If the coefficient of one of the variables in either equation is 1 or -1.
Elimination using addition
If one of the variables in the given two equations has the opposite coefficients (additive inverses). 
Elimination using subtraction
If one of the variables in the given two equations has the same coefficient.
Elimination using multiplication
If none of the variables has the coefficients 1 or -1 and if the variables cannot be eliminated by using addition or subtraction.

OCTOBER 21 CLASSWORK

oct 21 alg classwork

OCTOBER 21 CLASSWORK ANSWERS

 oct 21 alg answers

OCTOBER 22 IN CLASS PRACTICE TEST ANSWERS

oct 22 alg practice test answers

Practice Test FOR October 23

October 23 alg hw practice test

Practice Test ANSWERS FOR October 23

October 23 alg hw practice test answers

Elimination Method – Solving Systems of Equations

Let’s just do one and you’ll see how it works:

2x + 3y = 20 and -2x + y = 4
See how these guys are the same,
but with a different sign?

If we add the two equations — straight down, those x critters are going to drop right out!

Just add “like terms” and drag the “=” down to:

( 2x + 3y = 20 ) + ( -2x + y = 4 ) = ( 0 + 4y = 24 ) ... 4y = 24 ... y = 6

We’ve got one of them…  Now, we just need to get the x.  To do this, you can stick the y into either of the original equations…

The second equation is easier:

stick the y = 6 into -2x + y = 4 for y, which gives -2x + 6 = 4 ... -2x = -2 ... x = 1

It looks like the answer is (1, 6).

Solve this system of equations using the addition or subtraction method.  Check.

x – 2y = 14
x + 3y = 9

Most of the time they aren’t set up that nicely.  You’ll usually have to do a little dinking before something will drop out.

Look at this one:

3x - 4y = -5 and 5x - 2y = -6

If we just add straight down, nothing’s going to drop out and we’ll just get a mess.

But, check out the y guys:

3x - 4y = -5 and 5x - 2y = -6
If we could make this a +4y,
the
y‘s would drop out…

So, let’s do it!  Remember that we can multiply an equation by a number…  So, let’s multiply the second equation by a -2:

3x - 4y = -5 and -2 ( 5x - 2y = -6 )

Remember to hit each guy!

( 3x - 4y = -5 ) + ( -10x + 4y = 12 ) = ( -7x + 0 = 7 ) ... -7x = 7 ... x = -1

Now, stick the x guy into either of the original equations.  I’m going to go for the first one:

stick the x = -1 into 3x - 4y = -5 for x, which gives 3 ( -1 ) - 4y = -5 ... -3 - 4y = -5 ... -4y = -2 ... y = 1 / 2

The answer is (-1,1 / 2)

Let’s look at three more examples using the “addition” or “subtraction” method for systems of equations:

 

 

1.  Solve this system of equations
and check:
x – 2y = 14
x + 3y = 9
a.  First, be sure that the variables are “lined up” under one another.  In this problem, they are already “lined up”. x – 2y = 14
x + 3y = 9
b.  Decide which variable (“x” or “y“) will be easier to eliminate.  In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another.  Looks like “x” is the easier variable to eliminate in this problem since the x‘s already have the same coefficients. x – 2y = 14
x + 3y = 9
c.  Now, in this problem we need to subtract to eliminate the “x” variable.  Subtract ALL of the sets of lined up terms.
(Remember:  when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.)
 x – 2y = 14
 3y =  9

– 5y = 5  
d.  Solve this simple equation. -5y = 5
y =
 -1
e.  Plug “y = -1″ into either of the ORIGINAL equations to get the value for “x“.  x – 2y = 14
x – 2(-1) = 14
     x + 2 = 14
            x = 
12
f.  Check:  substitute x = 12 and y = -1 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE! x – 2y = 14
12
 – 2(-1) = 14
12 + 2 = 14
14 = 14  
(check!)
 + 3y = 9
12
 + 3(-1) = 9
12 – 3 = 9
9 = 9  
(check!)

 

2.  Solve this system of equations
and check:
4x + 3y = -1
5x + 4y = 1
a.  You can probably see the dilemma with this problem right away.  Neither of the variables have the same (or negative) coefficients to eliminate.  Yeek! 4x + 3= -1
5x + 4y = 1
b.  In this type of situation, we must MAKE the coefficients the same (or negatives)  by multiplication.   You can MAKE either the “x” or the “y” coefficients the same.  Pick the easier numbers.  In this problem, the “y” variables will be changed to the same coefficient by multiplying the top equation by 4 and the bottom equation by 3.
Remember:
* you can multiply the two differing coefficients to obtain the new coefficient if you cannot think of another smaller value that will work.
* multiply EVERY element in each equation by your adjustment numbers.  
4(4x + 3y = -1)
3
(5x + 4y = 1) 
16x + 12y = -4 15x + 12y = 3
c.  Now, in this problem we need to subtract to eliminate the “y” variable.
(
Remember:  when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.)
 16x + 12y = -4
15x  12y = – 3

         = – 7
d.  Plug “x = -7″ into either of the ORIGINAL equations to get the value for “y“.    5x + 4y = 1
5(-7) + 4y = 1
-35 + 4y = 1
4y = 36
               y = 9 
e.  Check:  substitute x = -7 and y = 9 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE! 4x + 3= -1
4(-7) +3(9) = -1
-28 + 27 = -1
-1 = -1  
(check!)   
5x + 4y = 1
5(-7) + 4(9) = 1
-35 + 36 = 1
1 = 1  
(check!)

 

 

Let’s finish with an addition method problem:

3.  Solve this system of equations and check: 4x – y = 10
2x  = 12 – 3y 
a.  First, be sure that the variables are “lined up” under one another.  The second equation was rearranged so that the variables would “line up” with those in the first equation. 4x  –   y = 10
2x + 3y = 12
b.  Decide which variable (“x” or “y“) will be easier to eliminate.  In this problem, we must MAKE EITHER the “x” or the “y” coefficients the same.  The “y” variable is being used here.  Multiplying by 3 will give the “y” variables negative coefficients.  (Yes, -3 could also have been used.)  3(4x – y = 10)
2x + 3y = 12  
12x – 3y = 30
2x + 3y = 12
c.  Now, add to eliminate the “y” variable.
12x – 3y = 30
2x + 3y = 12

14x        = 42 
d.  Solve this simple equation. 14x = 42
    x = 3
e.  Plug “= 3″ into either of the ORIGINAL equations to get the value for “y”.   4x – y = 10
4(3) – y = 10
12 – y = 10
y = -2
          y = 2
f.  Check:  substitute x = 3 and y = 2 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE! 4x – y = 10
4(3) – 2 = 10
12 – 2 = 10
10=10 
(check!)
2x  = 12 – 3y
2(3) = 12 – 3(2)
6 = 12 – 6
6 = 6  
(check!)

 

CLASSWORK FOR OCTOBER 16

Oct 16 alg classwork

Do Problems #1-6

 

 

OCTOBER 19 CLASSWORK 

Complete Problems #7-21 from Oct 16 worksheet

 OCTOBER 19 CLASSWORK ANSWERS

oct 19 alg classwork answers

 

OCTOBER 20 CLASSWORK

Do Problems #22-36

OCTOBER 20 CLASSWORK ANSWERS

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October 22 IN CLASS PRACTICE TEST ANSWERS

October 22 alg in class practice test answers (1)

Practice Test FOR October 23

October 23 alg hw practice test

Practice Test ANSWERS FOR October 23

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Substitution Method for Solving Linear Equations

The method of solving “by substitution” works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, “substituting” for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works.

Solve the following system by substitution.

2x – 3y = –2
4x +   y = 24

The idea here is to solve one of the equations for one of the variables, and plug this into the other equation. It does not matter which equation or which variable you pick. There is no right or wrong choice; the answer will be the same, regardless. But — some choices may be better than others.

For instance, in this case, can you see that it would probably be simplest to solve the second equation for “y =”, since there is already a y floating around loose in the middle there? I could solve the first equation for either variable, but I’d get fractions, and solving the second equation for x would also give me fractions. It wouldn’t be “wrong” to make a different choice, but it would probably be more difficult. Being smart, I’ll solve the second equation for y:

4x + y = 24 change the equation by solving for “y”
y = –4x + 24

Now I’ll plug this in (“substitute it”) for “y ” in the first equation, and solve for x:

2x – 3(–4x + 24) = –2
2x + 12x – 72 = –2
14x = 70
x = 5

Now I can plug this x-value back into either equation, and solve for y. But since I already have an expression for “y =”, it will be simplest to just plug into this:

y = –4(5) + 24 = –20 + 24 = 4

Then the solution is (x, y) = (5, 4).

 

 

 

CLASSWORK FOR October 14 

October 14 alg-classwork

 

CLASSWORK ANSWERS FOR October 14

October 14 alg-classwork-answers

 

 CLASSWORK FOR October 15

October 15 alg-sub-method-classwork

CLASSWORK ANSWERS FOR October 15

October 15 alg-sub-meth-answers

Graphing Systems of Linear Equations

What is a system of equations? 

A set of equations, for example, two equations with two unknowns, for which a common solution is sought is called a system of equations.

A MATH RIDDLE…  

Can you think of two numbers that when added together total 7, but when subtracted from each other their difference is 1? 

SOLUTION: 4 and 3 

This is just one example of a system of equations. Since the solution of a system must satisfy both conditions simultaneously, we say that we have a system of simultaneous equations.

Example:  Graph these equations

4x – 6y = 12
2x + 2y = 6

Put each equation in slope intercept form  and then graph them.

GrSysM1

GrSysM3

 

The point of intersection of the two lines, (3,0), is the solution to the system of equations.

graphtest1

This means that (3,0), when substituted into either equation, will make them both true.

 

 

 

Systems of equations can be solved algebraically or graphically. Usually, the problem is to find a solution for x and y that satisfies both equations simultaneously.

Graphically, this represents a point where the lines cross. There are 3 possible outcomes to this (shown here in blue, green, and red):



The two lines might not cross at all, as in

 

graph%28+300%2C+200%2C+-20%2C+20%2C+-20%2C+20%2C+x%2C+x%2B10%29+
y = x
y = x + 10.
This means there are no solutions, and the system is called inconsistent.

If you try to solve the system algebraically, you’ll end up with something that’s not true, such as 0 = 10.

Whenever you end up with something that’s not true, the system is inconsistent.


The two equations might actually be the same line, as in

 

graph%28+300%2C+200%2C+-20%2C+20%2C+-20%2C+20%2C+x%2B10%2C+x%2B10%29+
y = x + 10
2y = 2x + 20.
These are equivalent equations. The lines are actually the same line, and they ‘cross’ at infinitely many points (every point on the line). In this case, there are infinitely many solutions and the system is called dependent.

If you try to solve this system algebraically, you’ll end up with something that’s true, such as 0 = 0.

Whenever you end up with something that’s true, the system is dependent.



The two lines might cross at a single point, as in

 

graph%28+300%2C+200%2C+-20%2C+20%2C+-16%2C+24%2C+x%2B10%2C+2x%29+
y = x + 10
y = 2x.
If you try to solve this system algebraically, you’ll end up with something that involves one of the variables, such as x = 10. In this case, there is just one solution, and the system is called independent.

 

 Here are a couple of handy tables for recognizing what type of system you’re dealing with.

 

From the algebraic perspective:

 

 

If solving using the addition or substitution method leads to
then the system is
and the equations
X = a number, y = a number
independent
will have different values of m when both are placed in y = mx + b (slope-intercept) form
an inconsistent equation, such as 0 = 3
inconsistent
will have the same value of m, but different values of b, when both are placed in y = mx + b form
An identity, such as 5 = 5
dependent
will be identical when both are placed in slope-intercept form

From the graphical perspective:

 

 

If the equations have

then the system is

and the lines

Different slopes

independent

cross at a point

the same slope but different intercepts

inconsistent

are parallel and never cross

the same slope and the same intercept

dependent

are actually both the same line

 

 

 

OCTOBER 12 CLASSWORK 

October 12 alg classwork

October 12 alg classwork answers

OCTOBER 13 CLASSWORK 

Oct 13 alg classwork (2)

Oct 13 alg classwork answers (1)

 

October 14 alg hw   Do Problems #1-10

October 14 alg hw answers

 

 

 

STANDARD FORM : Graphing X and Y Intercepts

Quote

There are two efficient and accepted methods for graphing lines — this is one of them.  

On to method 1… First of all, you’ll need to know what an intercept is!

the graph of a line through the x-intercept and the y-intercept...the x-intercept is where the line goes through the x-axis, and the y-intercept is where the line passes through the y-axis

The intercepts are where a graph crosses the x and y axes.

Not only are these the easiest points to find, as you’ll see later in math, they are the most important points.

Here’s how this works… Let’s graph

3x - y = 6

Make an xy table:

a two column table with the left column for the x values of the two intercepts and the y column for the y values of the two intercepts

Fill in 0’s for the
x and y like this:

an xy table with 0 in the x axis for the x-intercept, and a 0 in the y column for the y-intercept

 

Let’s look at the first one:

by plugging x = 0 into the original equation 3x - y = 6, we get 3 (0) - y = 6, which simplifies to -y = 6 and gives y = -6

The easiest thing to do is to just cover up the 3x part with your finger:

ignore the 3x part of the equation 3x - y = 6 by covering it with your finger

So, we get

put the -6 in the y column next to the 0 in the x column, which shows that ( 0 , -6 ) is a point on our graph -- the y-intercept

 

Now, let’s look at the second one:

by plugging y = 0 into the original equation 3x - y = 6, we get 3x - 0 = 6, which simplifies to 3x = 6 and gives x = 2

Just cover the y guy with your finger:

ignore the -y part of the equation 3x - y = 6 by covering it with your finger

So, we get

put the 2 in the x column next to the 0 in the y column, which shows that ( 2 , 0 ) is a point on our graph -- the x-intercept

Since we just need two points to draw the line, we’re set!

the graph of a line passing through the points ( 0 , -6 ) and ( 2 , 0 )

Here’s another one:

Graph   

x + 2y = 4

1) start by making an xy table and putting a 0 in the x column and a 0 in the y column.  2) cover the x in the equation x + 2y = 4 with your finger, which shows y = 2


3) put the 2 in the y column next to the 0 in the x column.  4) cover the 2y in the equation x + 2y = 4 with your finger, which shows  x = 4


5) put the 4 in the x column next to the 0 in the y column.  6) graph a line using the two intercepts from the xy chart... a line through the points ( 0 , 2 ) and ( 4 , 0 )

EXAMPLE: 2x + y = -2
SOLUTION:
2x + y = -2
If x= 0, then y = -2
If y=0, then 2x= -2, so x= -1
Plot these points and draw the line through these points. It should look like this:

std form a

EXAMPLE:  3x – 6y= -3
SOLUTION:
3x  –  6y = -3
If x = 0, then  -6y =-3, so y= std form b
If y=0, then 3x = -3, so x = -1
Plot these points and draw the line connecting the points. It should look like this:

std form 1

 

 

 

CLASSWORK FOR  OCTOBER 8

oct 8 alg classwork

 

CLASSWORK ANSWERS  for OCTOBER 8

oct 8 alg classwork answers

Graphing Linear Equations in Slope Intercept Form


lines

These lines look pretty different, don’t they?

Lines are used to keep track of lots of info — like what is your GPA (grade point average) for all your classes..   Whether the line is tilted up or down, all of the sudden, gets REALLY important!

I am now going to introduce you to

Pierre The Mountain Climbing Ant

Pierre the Mountain Climbing Ant

For slopes, Pierre is going to walk on the lines from left to right
— just like we read.

On these lines, Pierre is climbing uphill.

Uphill slopes are positive slopes. That’s good because that means your GPA is going up
The slope will be a positive number like
5 or 2/3.

On these lines, Pierre is climbing downhill.

Downhill slopes are negative slopes. That means your GPA is going down.
The slope will be a negative number like
-7 or -1/3

 

slope intercept

 

I think the most useful form of straight-line equations is the “slope-intercept” form:

y = mx + b

This is called the slope-intercept form because

m” is the slope and

b” gives the y-intercept.

I like slope-intercept form the best. It is in the form “y=“, which makes it easiest to plug into, either for graphing or doing word problems. Just plug in your x-value; the equation is already solved for y.

But the best part about the slope-intercept form is that you can read off the slope and the intercept right from the equation. This is great for graphing.

y=5x+3 is an example of the Slope Intercept Form and represents the equation of a line with a slope of 5 and and a y-intercept of 3 .

y= −2x + 6 represents the equation of a line with a slope of −2 and and a y-intercept of 6

 

Let’s graph:

y = ( -3 / 5 )x + 4  ...  start with the + 4

 

1 ) The y-intercept is 4:

Graph of the point ( 0 , 4 )

2 ) The slope is -3 / 5 .  Starting from the y-intercept, move down three and right five

 

 Graph of a line that goes through the points ( 0 , 4 ) and ( 5 , 1 )  ...  go down 3  ... and over 5

 

The easiest way to graph such a line, is to plot the y-intercept first. Then, write the slope m in the form of a fraction, like rise over run, and from the y-intercept, count up (or down) for the rise, over (right or left) for the run, and put the next point. Then connect the two points and this is your line.

A couple of examples might be helpful.

EXAMPLE :       y = 3x + 2
SOLUTION:
Y-intercept = 2, slope =     plot-formula
Start by graphing the y-intercept by going up 2 units on the y-axis.
From this point go UP (rise) another 3 units, then 1 unit to the RIGHT (run), and put another point. This is the second point. Connect the points and it should look like this:
plot-formula1

 

EXAMPLE :      y = -3x + 2
SOLUTION:
Y-intercept = 2, slope =  -3
Start by graphing the y-intercept by going up 2 units on the y-axis.
From this point go DOWN (rise) 3 units, then 1 unit to the RIGHT (run), and put another point. This is the second point. Connect the points and it should look like this:
plot-formula2

EXAMPLE :        y =  plot-formula4  x + 2

SOLUTION:
Y-intercept = 2, slope =  plot-formula5
Start by graphing the y-intercept by going up 2 units on the y-axis.
From this point go UP (rise) another 3 units, then 5 unit to the RIGHT (run), and put another point. This is the second point. Connect the points and it should look like this:

plot-formula6

 

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